Solution 4.4:7b - Revision history

Tek at 07:44, 14 October 2008

Tek — Tue, 14 Oct 2008 07:44:15 GMT

←Older revision		Revision as of 07:44, 14 October 2008
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-	If we use the Pythagorean identity and write	+	If we use the Pythagorean identity and write <math>\sin^2\!x</math> as <math>1-\cos^2\!x</math>, the whole equation can be written in terms of <math>\cos x</math>,
-	<math>\sin ^{2}x</math>	+
-	as	+
-	<math>1-\cos ^{2}x</math>, the whole equation written in terms of	+
-	<math>\cos x</math>	+
-	~~becomes~~	+

-		+	{{Displayed math\|\|<math>2(1-\cos^2\!x) - 3\cos x = 0\,,</math>}}
-	<math>2~~\left~~( 1-\cos ^{2}x ~~\right~~)-3\cos x=0</math>	+
-		+
-		+
-	~~<math></math>~~	+

	or, in rearranged form,		or, in rearranged form,

		+	{{Displayed math\|\|<math>2\cos^2\!x + 3\cos x - 2 = 0\,\textrm{.}</math>}}

-	~~<math>2\cos ^{2}x+3\cos x-2=0</math>~~	+	With the equation expressed entirely in terms of <math>\cos x</math>, we can introduce a new unknown variable <math>t=\cos x</math> and solve the equation with respect to ''t''. Expressed in terms of ''t'', the equation is
-		+
-		+
-	With the equation expressed entirely in terms of	+
-	<math>\cos x</math>, we can introduce a new unknown variable	+
-	<math>t=\cos x</math>	+
-	and solve the equation with respect to	+
-	~~<math>~~t~~</math>~~. Expressed in terms of	+
-	~~<math>~~t~~</math>~~, the equation is	+
-		+
-		+
-	~~<math>2t^{2}+3t-2=0</math>~~	+
-		+

-	~~and this second-degree equation has the solutions~~	+	{{Displayed math\|\|<math>2t^2+3t-2 = 0</math>}}
-	~~<math>t=\frac~~{1}{~~2}</~~math>	+
-	~~and~~	+
-	<math>t=-2</math>	+
-	.	+

-	~~In terms of~~	+	and this quadratic equation has the solutions <math>t=\tfrac{1}{2}</math> and
-	~~<math>x</math>,~~ this ~~means that either~~	+	<math>t=-2\,</math>.
-	<math>~~\cos x~~=\~~frac~~{1}{2}</math>	+
-	or	+
-	<math>~~\text{cos }x~~=-\~~text{2}~~</math>. ~~The first case occurs when~~	+

		+	In terms of ''x'', this means that either <math>\cos x = \tfrac{1}{2}</math> or <math>\cos x = -2</math>. The first case occurs when

-	<math>x=\pm \frac{\pi }{3}+2n\pi </math>	+	{{Displayed math\|\|<math>x=\pm \frac{\pi}{3}+2n\pi\qquad</math>(''n'' is an arbitrary integer),}}
-	(	+
-	~~<math>~~n~~</math>~~	+
-	an arbitrary integer),	+

-	whilst the equation	+	whilst the equation <math>\cos x = -2</math> has no solutions at all (the values of cosine lie between -1 and 1).
-	<math>\~~text{~~cos }x=-~~\text{~~2 }</math>	+
-	has no solutions at all (the values of cosine lie between	+
-	~~<math>~~-~~\text{~~1 ~~}</math>~~	+
-	and	+
-	~~<math>\text{~~1~~}</math>~~	+
-	).	+

	The answer is that the equation has the solutions		The answer is that the equation has the solutions

		+	{{Displayed math\|\|<math>x = \pm\frac{\pi}{3} + 2n\pi\,,</math>}}

-	~~<math>x=\pm \frac{\pi }{3}+2n\pi </math>~~	+	where ''n'' is an arbitrary integer.
-	(	+
-	~~<math>~~n~~</math>~~	+
-	an arbitrary integer).	+

Ian at 13:01, 1 October 2008

Ian — Wed, 01 Oct 2008 13:01:26 GMT

←Older revision		Revision as of 13:01, 1 October 2008
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-	{~~{NAVCONTENT_START}~~}	+	If we use the Pythagorean identity and write
-	<~~center~~> ~~[[Image:4_4_7b-~~1(2).~~gif]]~~ </~~center~~>	+	<math>\sin ^{2}x</math>
-	{{~~NAVCONTENT_STOP~~}}	+	as
-	{{~~NAVCONTENT_START}~~}	+	<math>1-\cos ^{2}x</math>, the whole equation written in terms of
-	<~~center~~> ~~[[Image:4_4_7b~~-2(2)~~.gif]]~~ </~~center~~>	+	<math>\cos x</math>
-	{{~~NAVCONTENT_STOP~~}}	+	becomes
		+
		+
		+	<math>2\left( 1-\cos ^{2}x \right)-3\cos x=0</math>
		+
		+
		+	<math></math>
		+
		+	or, in rearranged form,
		+
		+
		+	<math>2\cos ^{2}x+3\cos x-2=0</math>
		+
		+
		+	With the equation expressed entirely in terms of
		+	<math>\cos x</math>, we can introduce a new unknown variable
		+	<math>t=\cos x</math>
		+	and solve the equation with respect to
		+	<math>t</math>. Expressed in terms of
		+	<math>t</math>, the equation is
		+
		+
		+	<math>2t^{2}+3t-2=0</math>
		+
		+
		+	and this second-degree equation has the solutions
		+	<math>t=\frac{1}{2}</math>
		+	and
		+	<math>t=-2</math>
		+	.
		+
		+	In terms of
		+	<math>x</math>, this means that either
		+	<math>\cos x=\frac{1}{2}</math>
		+	or
		+	<math>\text{cos }x=-\text{2}</math>. The first case occurs when
		+
		+
		+	<math>x=\pm \frac{\pi }{3}+2n\pi </math>
		+	(
		+	<math>n</math>
		+	an arbitrary integer),
		+
		+	whilst the equation
		+	<math>\text{cos }x=-\text{2 }</math>
		+	has no solutions at all (the values of cosine lie between
		+	<math>-\text{1 }</math>
		+	and
		+	<math>\text{1}</math>
		+	).
		+
		+	The answer is that the equation has the solutions
		+
		+
		+	<math>x=\pm \frac{\pi }{3}+2n\pi </math>
		+	(
		+	<math>n</math>
		+	an arbitrary integer).

Tekbot: Lösning 4.4:7b moved to Solution 4.4:7b: Robot: moved page

Tekbot — Tue, 09 Sep 2008 10:08:03 GMT

Lösning 4.4:7b moved to Solution 4.4:7b: Robot: moved page

←Older revision

Revision as of 10:08, 9 September 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

Tekbot — Thu, 21 Aug 2008 07:21:09 GMT

Robot: Automated text replacement (-[[Bild: +[[Image:)

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Oskar — Thu, 03 Apr 2008 12:40:40 GMT

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